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An element of a
set is a
unit if its
multiplicative inverse belongs to that set. (The set of units form a multiplicative group.)
Algebraic integer units have
complex norm and are thus complex
roots of unity.
Units in quadratic integer rings
Imaginary
quadratic integer rings have only a few units (see
Theorem I1). The
primitive root of unity of degree
 $\omega :=e^{i{\tfrac {2\pi }{3}}}={\frac {1}{2}}+{\frac {\sqrt {3}}{2}},$
is used in the following table.
Units of imaginary quadratic integer rings

Units

@opsp@@opsp@≤@opsp@@opsp@ @opsp@−@opsp@5

$\{(1)^{0},(1)^{1}\}=\{1,1\}$

@opsp@−@opsp@3

$\{\omega ,\,\omega ,\omega ^{2},\,\omega ^{2},1,1\}$

@opsp@−@opsp@2

$\{(1)^{0},(1)^{1}\}=\{1,1\}$

@opsp@−@opsp@1

$\{i^{0},i^{1},i^{2},i^{3}\}=\{1,i,1,i\}$

Real quadratic integer rings have infinitely many units (see
Theorem R1), which are all powers of each other, some multiplied by
. For that reason, the table below can't be complete like the table above.
An inefficient way to find a unit of a real quadratic integer ring
other than
or
is to try each positive value of
starting with
and going up until
is an integer.
Units of real quadratic integer rings

Units

2

$1+{\sqrt {2}}$

3

$2+{\sqrt {3}}$

4

N/A

5

${\frac {1}{2}}+{\frac {\sqrt {5}}{2}}$ (golden ratio)

6

$5+2{\sqrt {6}}$

7

$8+3{\sqrt {7}}$

8

N/A, but note that $3^{2}8(1^{2})=1$

9

N/A

10

$3+{\sqrt {10}}$

11

$10+3{\sqrt {11}}$

12

N/A, but $7^{2}12(2^{2})=1$

13

${\frac {3}{2}}+{\frac {\sqrt {13}}{2}}$

14

$15+4{\sqrt {14}}$

15

$4+{\sqrt {15}}$

16

N/A

17

$4+{\sqrt {17}}$

18

N/A, but $17^{2}18(4^{2})=1$

19

$170+39{\sqrt {19}}$

20

N/A, but $9^{2}20(2^{2})=1$

21

${\frac {5}{2}}+{\frac {\sqrt {21}}{2}}$

22

$197+42{\sqrt {22}}$

23

$24+5{\sqrt {23}}$

24

N/A, but $5^{2}24(1^{2})=1$

25

N/A

26

$5+{\sqrt {26}}$

27

N/A, but $26^{2}27(5^{2})=1$

28

N/A, but $127^{2}28(24^{2})=1$

29

${\frac {5}{2}}+{\frac {\sqrt {29}}{2}}$

30

$11+2{\sqrt {30}}$

31

$1520+273{\sqrt {31}}$

32

N/A, but $17^{2}32(3^{2})=1$

33

${\frac {23}{2}}+{\frac {4{\sqrt {33}}}{2}}$

Examples
See also